poj_1035

源链接:http://poj.org/problem?id=1035

水题,只要读清楚题目,按照题目要求做即可。

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-8
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll __int64
int n,m;
#define Mod 1000000007
#define N 110
#define M 10010
char dic[M][25];
char word[55][25];
map<string,int> mp;
bool check(char *s,char *p){
	int len1 = strlen(s);
	int len2 = strlen(p);
	if(len1 != len2 && abs(len1-len2)!=1)
		return false;
	int dif = 0;
	if(len1 == len2){
		for(int i=0;i<len1;i++)
			if(s[i]!=p[i])
				dif++;
		return (dif==1);
	}else if(len1>len2){
		int i,j;
		for(i=0,j=0;i<len1&&j<len2;){
			if(s[i] == p[j]){
				i++;
				j++;
			}else{
				i++;
			}
		}
		return (j==len2);
	}else if(len1<len2){
		int i,j;
		for(i=0,j=0;i<len1&&j<len2;){
			if(s[i] == p[j]){
				i++;
				j++;
			}else{
				j++;
			}
		}
		return (i==len1);
	}
}
void deal(char *s){
	if(mp[s]){
		printf("%s is correct\n",s);
		return;
	}
	printf("%s:",s);
	for(int i=0;i<n;i++){
		if(check(dic[i],s)){
			printf(" %s",dic[i]);
		}
	}
	printf("\n");
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//  freopen("out.txt","w",stdout);
#endif
    	while(sfs(dic[0])!=EOF){
    		int p1 = 1;
    		mp.clear();
    		mp[dic[0]] = 1;
    		while(sfs(dic[p1])){
    			if(strcmp(dic[p1],"#") == 0)
    				break;
    			mp[dic[p1]] = 1;
    			p1++;
    		}
    		int p2 = 0;
    		while(sfs(word[p2])){
    			if(strcmp(word[p2],"#") == 0)
    				break;
    			p2++;
    		}
    		n = p1;
    		for(int i=0;i<p2;i++){
    			deal(word[i]);
    		}
    	}
return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值